[Zope] Trying to compile ZPyGresql

Tom Deprez tom.deprez@uz.kuleuven.ac.be
Mon, 14 Feb 2000 12:11:41 +0100


Can somebody help me on this topic : eg installing the ZPyGresql?

When I try to compile the module I get the following output

rm -f *.o *~
rm -f *.a tags TAGS config.c Makefile.pre python sedscript
rm
-f *.so *.sl so_locations
VERSION=`python -c "import sys; print
sys.version[:3]"`; \
installdir=`python -c "import sys; print sys.prefix"`;
\
exec_installdir=`python -c "import sys; print sys.exec_prefix"`; \
make
-f ./Makefile.pre.in VPATH=. srcdir=. \
	VERSION=$VERSION \

installdir=$installdir \
	exec_installdir=$exec_installdir \

Makefile
make[1]: Entering directory
`/usr/local/zope/lib/python/Products/ZPyGreSQLDA/src'
make[1]: *** no rule
to make target '/usr/lib/python1.5/config/Makefile', needed by 'sedscript'.
Stop.
make[1]: Leaving directory
`/usr/local/zope/lib/python/Products/ZPyGreSQLDA/src'
make: ***[boot] Error2

What does this error-line mean: "no rule to make target..."

If I know what this line means, then I'm already a little bit further in
finding a solution to this error.

Is this the best way to connect zope with PostGressql? 
ODBC better? Is installing an ODBC on linux the same as on Windows?
Where can I find good ODBC to work with Zope and PostGreSQL on Linux?

Thanks in advance,

Tom.